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Grade: 11
        If graph of expression y= x^2 - 8x + 12 is shown in figure (attached IMG) Calculate area of square ABCD inscribed between parabola and x-axis
9 months ago

Answers : (2)

Ravindranath Jada
12 Points
							Consider the points of square as (a,0)(b,0)(a,c) and (b,c)Side of the square is b-a (horizontally)if b>a and c (vertically)c=b-a ---(1)Also the points (a,c) and (b,c) lie on the parabolac=a^2-8a+12 ----(2)c= b^2-8b+12 ---(3)Subtracting 2 and 3 we geta^2-b^2-8a+8b=0(a-b)(a+b)=8(a-b)a+b=8 ---(4)b=8-aSolve equation 1 and 3 using equation 4c=b-(8-b)=2b-82b-8=b^2-8b+12b^2-10b+20=0(b-5)^2=5b-5=√5,-√5b=5+√5,5-√5But the curve intersects x-axis at x=2 and x=6So value of b is 5-√5Value of a is 8-(5-√5)=3+√5Side of the square is |b-a| = 2√5Area of the square =( √20 )^2 = 20 sq.units
						
9 months ago
Yasas Chandra Sai P
17 Points
							Consider the points of square as (a,0)(b,0)(a,c) and (b,c)Side of the square is b-a (horizontally)if b>a and c (vertically)|c|=|b-a| ---(1)Also the points (a,c) and (b,c) lie on the parabolac=a^2-8a+12 ----(2)c= b^2-8b+12 ---(3)Subtracting 2 and 3 we geta^2-b^2-8a+8b=0(a-b)(a+b)=8(a-b)a+b=8 ---(4)b=8-aSolve equation 1 and 3 using equation 4|c|=|b-(8-b)|=|2b-8|2b-8 = -(b^2-8b+12)b^2-6b+4=0(b-3)^2=5b-3=√5,-√5b=3+√5,3-√5But the curve intersects x-axis at x=2 and x=6So value of b is 3+√5Value of a is 8-(3+√5)=5-√5Side of the square is |b-a| = 2√5-2Area of the square =( b-a )^2 = (2√5-2)^2 sq.units
						
9 months ago
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